Differential Equations Summary: First and Second Order Equations, Modeling Problems, Study notes of Differential Equations

A summary of first and second-order differential equations and how to solve them. It also explains how to model a problem using differential equations. The first-order DEs are solved using variables separable DE and reduction through substitution. The second-order DEs are solved by integrating twice. The document also provides an example of how to model a problem using differential equations.

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Differential Equations Summary
1. First order differential equations
a. Variables Separable DE:
Arrange through manipulation such that the form below is achieved:
dyygdxxf )()(
=
Integrate subsequently to yield the required solution.
Example: Solve
y
dx
dy = 1
for y<1.
SOLUTIO :
1
1
1
1=
= dx
dy
y
y
dx
dy
=
dxdy
y1
1
Cxy
+
=
|1|ln
Since
,1
<
y
(
)
Cxy += 1ln
Bx
ey
+
=1
x
Aey
= 1 where
B
eA
=
,
cB
=
(shown)
This solution is commonly termed the GEERAL SOLUTIO, where
A
is
unknown. When initial conditions are provided, eg
y
=0 when
x
=0, then
A
assumes a specific value and the solution is termed the PARTICULAR SOLUTIO.
When we use the GC to plot out a series of graphs for various values of
A
, the result
is that we produce a family of solution curves.
b. Reduction through substitution:
The introduction of an intermediate variable aids in reducing the original
differential equation to a far simpler version which is readily solvable.
Example: Use the substitution y=
, where
v
is a function of x, to solve the
differential equation yx
dx
dy
x
+=
3
.
SOLUTIO :
dx
dv
xv
dx
dy
vxy
+==
Substituting into the differential equation gives
vxx
dx
dv
xvx +=
+3
pf3

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Differential Equations Summary

1. First order differential equations

a. Variables Separable DE:

Arrange through manipulation such that the form below is achieved:

f ( x)dx=g(y) dy

Integrate subsequently to yield the required solution.

Example: Solve y dx

dy = 1 − for y<1.

SOLUTIO :

dx

dy

y

y dx

dy

− dy dx 1 y

− ln| 1 −y |=x+ C

Since y < 1 , − ln( 1 −y) =x+C

x B y e

−+ 1 − =

x y Ae

− = 1 − where

B A = e , B = −c (shown)

This solution is commonly termed the GEERAL SOLUTIO , where A is

unknown. When initial conditions are provided, eg y =0 when x =0, then A

assumes a specific value and the solution is termed the PARTICULAR SOLUTIO.

When we use the GC to plot out a series of graphs for various values of A , the result

is that we produce a family of solution curves.

b. Reduction through substitution:

The introduction of an intermediate variable aids in reducing the original

differential equation to a far simpler version which is readily solvable.

Example: Use the substitution y= vx , where v is a function of x, to solve the

differential equation x y dx

dy x = 3 +.

SOLUTIO :

dx

dv v x dx

dy y =vx⇒ = +

Substituting into the differential equation gives

x vx dx

dv x v x = +

dx x

dv x dx

dv x

2 = ⇒ =

= dx x

dv

∴ v = 3 ln|x|+C

y = 3 x ln|x|+ Cx (shown)

2. Second order differential equations:

Second order DEs are typically of the form ( ) 2

2

f x dx

d y = , whereby running the

DE through two iterations of integration will yield the required solution.

Example: g dt

d s =− 2

2

Integrating twice wrt t gives: gt A dt

ds =− +

and At B

gt s = − + + 2

2

(shown)

3. Modelling a problem through the usage of a differential equation:

Typically the question demands the student to first formulate a DE relating

to the context of the situation, and subsequently solve it. Realise that the

formulation of a DE involves the following considerations:

(i) Constants of proportionality

(ii) Net rate, which is usually composed of an “in” rate and “out” rate,

eg birth rate - death rate

Example: The growth of a particular species of insect is studied in an experimental

environment. The rate of death is proportional to the number in

thousands, x , of insects, at any time t days after the start of the

experiment. The rate of birth is proportional to.

2 x When x= 2 ,

the number of larvae hatched is equal to the number of insects that

died. Show that = ax( x− 2 ) dt

dx where a is a constant.

SOLUTIO :

dt

dx Birth rate minus death rate

= ax −bx

2

When x = 2 , = dt

dx 0